Beautiful problem on a progression

$$\require{cancel}\begin{align} x_{n+2}&=x_n-\frac 1{x_{n+1}}\\ x_{n+1}x_{n+2}&=x_n x_{n+1}-1\\ \end{align}$$

Given that $x_1=20$ and $x_2=14$, we have $$\begin{align} x_1 x_2&=280\\ x_2 x_3&=279\\ x_3 x_4&=278\\ \vdots &= \vdots\\ x_{280} x_{281}&=1\\ x_{281} x_{282}&=0\\ \end{align}$$

Hence $$x_{282}=0$$

i.e. $x_n=0$ when $n=282$.

NB - technically the series is not defined for $n>282$ because $x_{283}$ has a $\frac 10$ term; however if it is accepted that $\frac 10=\infty$ and $\frac 1\infty=0$, then every even term after the $282^{\text{nd}}$ is zero. An interesting problem though.

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Further to my solution above, this is in response to the request for expressing $x_n$ explicitly.

We can see that $$\begin{align} x_1=20&, x_2=14\\ x_3=\frac {279}{14}&, x_4=14\cdot \frac{278}{279}\\ x_5=\frac{277}{278}\cdot \frac{279}{14}&, x_6=14\cdot \frac{278\cdot 276}{279\cdot 277}\\ x_7=\frac{275\cdot 277}{276\cdot 278}\cdot \frac{279}{14}&, x_8=14\cdot \frac{278\cdot 276\cdot 274}{279\cdot 277\cdot 275}\\ \vdots &,\vdots \end{align}$$ hence, $x_n$ can be explicitly defined depending on whether $n$ is odd or even, i.e. $$x_{2m}=14\cdot \frac{\prod_{r=0}^{m-2}278-2r}{\prod_{r=0}^{m-2}279-2r}= 14\cdot \frac{278^\underline{\underline{m-1}}}{279^\underline{\underline{m-1}}} =\frac 1{20}\cdot \frac{280^\underline{\underline{m}}\;\;\;}{279^\underline{\underline{m-1}}}$$ and $$x_{2m+1}=\frac{279}{14}\cdot \frac{\prod_{r=0}^{m-2}277-2r}{\prod_{r=0}^{m-2}278-2r} =\frac 1{14}\cdot \frac{279^\underline{\underline{m}}\;\;\;}{278^\underline{\underline{m-1}}} =20\cdot \frac{279^\underline{\underline{m}}}{280^\underline{\underline{m}}}$$ where $m\geq 2$, and $x^\underline{\underline{n}}=\overbrace{x(x-2)(x-4)\cdots (x-2n+2)}^{n \text{ terms}}$

Hint: $x_{n+2} x_{n+1} = x_{n+1} x_{n} - 1$.